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Impulse Momentum Method for Rotational

  1. Aug 6, 2012 #1
    1. The problem statement, all variables and given/known data
    I have abit of problem understanding this 2 question, I would appreciate if someone can guide me along.

    For diagram 1 (Left)
    I interpret the condition as: At 0sec, initial velocity is 0, At 5sec final velocity is unknown.


    For diagram 2 (Right)
    I interpret the condition as: At 0 sec, initial velocity is 0, At 3sec final velocity is 14.5m/s


    Please let me know if I get any of the condition wrong, thank you.
     

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  2. jcsd
  3. Aug 6, 2012 #2
    I think you understand the conditions correctly.
     
  4. Aug 6, 2012 #3
    o_0 hmm strange then how come i cannot get the answer...
     
  5. Aug 6, 2012 #4
    What goes wrong?
     
  6. Aug 6, 2012 #5
    For Diagram 1 part A

    M1=mv
    =0

    M2=mv
    =(20)(V2)
    =20V2

    M1+I1-2=M2
    981-5TA=20V2
    5TA=981-20V2
    TA=196.2-4V2

    Ans is TA=196.2-4V

    4V2 and 4V are both different meaning right?
     
  7. Aug 6, 2012 #6
    Why are you ignoring the tangential force?
     
  8. Aug 6, 2012 #7
    You mean the 75N force?

    What has it got to do with part A?
     
    Last edited: Aug 6, 2012
  9. Aug 6, 2012 #8
    You have already solved part A as far as I can tell.
     
  10. Aug 6, 2012 #9
    My Answer is: TA=196.2-4V2
    Question Ans is : TA=196.2-4V

    Is there any difference between V2 and V? one is initial one is final.
     
  11. Aug 6, 2012 #10
    The initial velocity is zero per the conditions ("released at t = 0"). So v can only be the final.
     
  12. Aug 6, 2012 #11
    How about the subscript? Don't have to write?
     
  13. Aug 6, 2012 #12
    Up to you. You can say "let's denote the velocity at t = 5 s as V" or in any other way. That does not change the underlying physics. You just need to know what denotes what.
     
  14. Aug 6, 2012 #13
    Part B

    M1=Iω
    =0

    M2=Iω
    =(2.5)(V/0.15)
    =16.67V

    I1-2=Torque x Time
    =[-22.5+Ta(0.15)]5
    =34.65-3V

    M1+I1-2=M2
    34.65-3V=16.67V
    34.65=19.67V
    V=1.76

    V=rω
    ω=1.76/0.15
    ω=11.74 - Ans Wrong
     
    Last edited: Aug 6, 2012
  15. Aug 6, 2012 #14
    Hmm. I have used a somewhat different method, but got the same result. Let's see if somebody else here can point out our mistake.
     
  16. Aug 7, 2012 #15

    Doc Al

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    Staff: Mentor

    OK.

    Where did you get the 22.5 from? Redo that calculation.
     
  17. Aug 7, 2012 #16
    Hi Doc Al, thanks for helping.

    I got my -22.5 from the tangential force Ft=75N I used the 75 times 0.3 = 22.5 Because Tension x Radius = Torque


    *Ok i think i got it, it's 300mm, 150mm diameters
     
  18. Aug 7, 2012 #17

    Doc Al

    User Avatar

    Staff: Mentor

    There you go. :wink:
     
  19. Aug 7, 2012 #18
    M1=Iω
    =0

    M2=Iω
    =(2.5)(V/0.15)
    =16.67V

    I1-2= Torque x Time
    =(TaX0.075-75X0.15)5
    =17.325-1.5V

    M1+I1-2=M2
    17.325-1.5v=16.67v
    17.325=18.17v
    v=0.95

    v=rω
    0.95=0.15ω
    ω=6.3 wrong again :cry:
     
  20. Aug 7, 2012 #19

    Doc Al

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    Staff: Mentor

    Same problem with radius vs diameter. (Sorry I didn't spot it before.)
     
  21. Aug 7, 2012 #20
    how come i need to take the radius of the axle and not the wheel?
     
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